Ask question

Ask question

All categories

2 years ago

9

Anumeha is mowing lawns for a summer job. For every mowing job, she charges an initial fee of $10 plus a constant fee for each h

our of work. Her fee for a 5-hour job, for instance, is $35.
Let F(t) denote Anumeha's fee for a single job F (measured in dollars) as a function of the number of hours "t" it took her to complete it.
Write the function's formula.

2 answers:

Andrew [12]2 years ago

8 0

Answer:

$F(t)=10+7t$

Step-by-step explanation:

We know that Anumeha charges an initial fee of $10, this is an initial condition which represents the independent term in the function.

Then, she charges $35 for 5 hours, this means that the ratio of change is

$\frac{35}{5}=7$

That is, she charges $7 per hour. Now, let's call $t$ the hours, the given problem is modelled with the following expression

$F(t)=10+7t$

Therefore, the function's formula is

$F(t)=10+7t$

timurjin [86]2 years ago

5 0

F(t)=5t+10, 5 dollars for each our plus 10 dollar initial fee

You might be interested in

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

2 years ago

What is next number 3469131824

-BARSIC- [3]

The next number would be 3469131825 you just change the last number
hope it helps

7 0

2 years ago

Read 2 more answers

If a geometric sequence has a1 = 256 and a8 = 512, find the exact value of the common ratio ????.

marusya05 [52]

Answer:r=1.104

Step-by-step explanation:

Given

First term a_1=256

$a_8=512$

Also $a_8$ can also be written in

$a_8=ar^7$

where r=common difference

$512=256r^7$

$2=r^7$

$r=\left ( 2\right )^{\frac{1}{7}}$

r=1.104

7 0

2 years ago

A town has 10,000 two-child families. Design a simulation to estimate the percentage of two-child families with two girls. Choos

shepuryov [24]

Given is a town has 10,000 two-child families. The town should be divided by smaller units and then conduct a survey on these units to account for the number of two-child families per unit. Combine the number per unit and divide by the total population of the town. Construct a table containing the name of the units and the number of two-child family per unit.

6 1

2 years ago

Read 2 more answers

B. if lucinda has only $18 to spend and the price of kewpie dolls and the price of beanie babies are both $6, how many of each w

adoni [48]

Given the table below comparing the marginal benefit Lucinda gets from Kewpie dolls and Beanie Babies.

$\begin{tabular}
{|p {2cm}|p {2cm}|p {2cm}|p {2cm}|}
\multicolumn {4} {|c|} {Lucinda's Kewpie Doll and Beanie Baby Marginal Benefits}\\[1ex]
\multicolumn {2} {|c|} {Kewpie Dolls}&\multicolumn {2} {|c|} {Beanie Babies}\\[1ex]
1&\$15.00&1&\$12.00\\
2&\$12.00&2&\$10.00\\
3&\$9.00&3&\$8.00\\
4&\$6.00&4&\$6.00\\
5&\$3.00&5&\$4.00\\
6&\$0.00&6&\$2.00\\
\end{tabular}$

<span>If lucinda has only $18 to spend and the price of kewpie dolls and the price of beanie babies are both $6,

Lucinda will buy the combination for which marginal benefit is the same.

Therefore, Lucinda will buy </span><span>2 kewpie dolls and 1 beanie baby,</span><span> if she were rational.</span>

5 0

2 years ago